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Electrostatics question

2005 · Q172
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Electrostatics question

2005 · Q172

NEETPhysicsElectrostaticsMCQ+4 / −1
Two charges q1 and q2 are placed 30 cm apart, as shown in the figure. A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D.

The change in the potential energy of the system is q34πε0{{{q_3}} \over {4\pi {\varepsilon _0}}}4πε0​q3​​ where k is

AIPMT 2005 Physics - Electrostatics Question 38 English
  1. A
    8q1
  2. B
    6q1
  3. C
    8q2
  4. D
    6q2
View written solutionFree

Correct answer: C

We know that potential energy of discrete system of charges is given by

U=14πε0(q1q2r12+q2q3r23+q3q1r31)U = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {{r_{12}}}} + {{{q_2}{q_3}} \over {{r_{23}}}} + {{{q_3}{q_1}} \over {{r_{31}}}}} \right)U=4πε0​1​(r12​q1​q2​​+r23​q2​q3​​+r31​q3​q1​​)

According to question,

Uinitial=14πε0(q1q20.3+q2q30.5+q3q10.4){U_{initial}} = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {0.3}} + {{{q_2}{q_3}} \over {0.5}} + {{{q_3}{q_1}} \over {0.4}}} \right)Uinitial​=4πε0​1​(0.3q1​q2​​+0.5q2​q3​​+0.4q3​q1​​)

Ufinal=14πε0(q1q20.3+q2q30.1+q3q10.4){U_{final}} = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {0.3}} + {{{q_2}{q_3}} \over {0.1}} + {{{q_3}{q_1}} \over {0.4}}} \right)Ufinal​=4πε0​1​(0.3q1​q2​​+0.1q2​q3​​+0.4q3​q1​​)

Ufinal−Uinitial=14πε0(q1q20.1−q2q30.5){U_{final}} - {U_{initial}} = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {0.1}} - {{{q_2}{q_3}} \over {0.5}}} \right)Ufinal​−Uinitial​=4πε0​1​(0.1q1​q2​​−0.5q2​q3​​)

=14πε0[10q2q3−2q2q3]=q34πε0(8q2) = {1 \over {4\pi {\varepsilon _0}}}\left[ {10{q_2}{q_3} - 2{q_2}{q_3}} \right] = {{{q_3}} \over {4\pi {\varepsilon _0}}}\left( {8{q_2}} \right)=4πε0​1​[10q2​q3​−2q2​q3​]=4πε0​q3​​(8q2​)

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